Widgets of Type A arrive with Poisson Process with arrival rate $\lambda_A$, and, for Type B, with arrival rate $\lambda_B$ (independent).

During t, there have been b arrivals of Type B. What are the expected arrivals of Type A+B in time frame t?

Does one simply take the given value b, and add to that the expected arrivals for process A?:

$b+t \times \lambda_A$

  • 2
    $\begingroup$ Assuming that both processes are independent, you can do like that. otherwise, it is inconclusive unless some auxiliary conditions are assumed. $\endgroup$ – Sangchul Lee Oct 22 '12 at 6:56
  • $\begingroup$ thanks but why do you not post this as an answer, but as a comment? $\endgroup$ – Wuschelbeutel Kartoffelhuhn Oct 22 '12 at 7:00
  • $\begingroup$ I usually post a comment when I think the answer lacks the required details to be an answer... $\endgroup$ – Sangchul Lee Oct 22 '12 at 7:39
  • $\begingroup$ Wuschel: You could transform this into an acceptable question by adding the information @sos440 suggested. $\endgroup$ – Did Oct 22 '12 at 13:58
  • $\begingroup$ @did Ok i added the indep. assumption $\endgroup$ – Wuschelbeutel Kartoffelhuhn Oct 23 '12 at 4:16

In general, $N_t=N^A_t+N^B_t$ implies $\mathbb E(N_t\mid N^B_t=b)=\mathbb E(N^A_t\mid N^B_t=b)+b$. If furthermore the processes $N^A$ and $N^B$ are independent, then $\mathbb E(N^A_t\mid N^B_t=b)=\mathbb E(N^A_t)=\lambda_At$.

Thus, in your setting, $\mathbb E(N_t\mid N^B_t=b)=b+\lambda_At$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.